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Use the image and table to answer the question.

An illustration shows a scatterplot titled Number of Laps the Swim Team Completes Per Week. The x-axis shows weeks ranging from 1 through 10 in increments of one. The y-axis shows number of laps and ranges from 100 to 1,000 in 100 unit increments.

Week Number of Laps
1 50
2 150
3 200
4 300
5 500
6 600
7 700
8 800
Which of the following is an appropriate regression model, rounded to the nearest tenth, given the bivariate dataset?

(1 point)
Responses

f(x)∼111.9x+50
f left parenthesis x right parenthesis is similar to 111.9 x plus 50

f(x)∼−111.9x−91.1
f left parenthesis x right parenthesis is similar to negative 111.9 x minus 91.1

f(x)∼111.9x
f left parenthesis x right parenthesis is similar to 111.9 x

f(x)∼111.9x−91.1

1 answer

To determine the appropriate regression model for the given bivariate dataset based on the scatterplot and data provided, we need to analyze how the number of laps completed changes with respect to the number of weeks.

Given the data points from the table:

  • Week 1: 50 laps
  • Week 2: 150 laps
  • Week 3: 200 laps
  • Week 4: 300 laps
  • Week 5: 500 laps
  • Week 6: 600 laps
  • Week 7: 700 laps
  • Week 8: 800 laps

We can note that the number of laps generally increases as the number of weeks increases.

Using a method such as least squares regression, you would typically calculate the best fit line for the data. The response options provided include different forms of linear equations and their coefficients.

Let's analyze the regression options:

  1. f(x)∼111.9x+50
  2. f(x)∼−111.9x−91.1
  3. f(x)∼111.9x
  4. f(x)∼111.9x−91.1

Given the general upward trend observed in the dataset, options with a negative slope (like Option 2) would not be appropriate.

Among the options with a positive slope, we should consider the y-intercept:

  • In Option 1, the intercept is 50 which might fit the data.
  • In Option 3, the linear relation starts at the origin which doesn't fit the data at Week 1.
  • In Option 4, the intercept of -91.1 might indicate a line that does not accurately reflect the observed values.

We want the model that best fits the dataset. Without detailed calculation or plotting, but based on inspection of the values, f(x)∼111.9x+50 appears to be a reasonable choice as it captures both the slope and an appropriate starting point, representing the progression from Week 1 correctly.

Thus, the final answer is:

f(x)∼111.9x+50